, 2005

Solution:

a. Consider the four corners and the two midpoints of the sides of length $6$. The distance between any two of these six points is $3$ or more, so one circle cannot cover two of these points, and at least six circles are needed.

On the other hand, one circle will cover a $2 \times 2$ square, and it is easy to see that six such squares can cover the rectangle.

b. Consider the four corners and the centre of the rectangle. The minimum distance between any two of these points is the distance between the centre and one of the corners, which is $\sqrt{34}/2$. This is greater than the diameter of the circle $\left(\sqrt{34/4} > \sqrt{32/4}\right)$, so one circle cannot cover two of these points, and at least five circles are needed.

Partition the rectangle into three rectangles of size $5/3 \times 2$ and two rectangles of size $5/2 \times 1$ as shown on the right. It is easy to check that each has a diagonal of length less than $2\sqrt{2}$, so five circles can cover the five small rectangles and hence the $5 \times 3$ rectangle.